How to solve this half-life problem

A fossil is found to have a 14C level of 63.0% compared to living organisms. How old is the fossil?

given: half-life of 14C = 5730 years

ln(Nt/No)= -0.693 t/ t half
ln(0.63) = -0.693(t)/ 5730
-0.462= -0.693 t / 5730
-2647.46 = -0.693 t
3820.29 years old =t

e^(t*(-ln(2)/HL))=.63 ---> ln(.63)=t(-ln(2)/HL) ---> t = -HL*ln(.63)/ln(2) ... t is the age of the fossil .. HL(half-life) .. ln (natural logarithm )

m = mo/2^x

0.63 = 1 / 2^x

x = ln1.59 / ln2
x = .67
ans: 0.67 x 5730
=3.833